The mangnitude of pontential energy of electron in `n^(th)` excited state of `He^(+)` ion is `(8)/(81)` time the kinetic energy of electron of excited state of `Li^(2+)` ion. Find `'n'`

Explanation

To solve the problem step by step, we will analyze the potential and kinetic energy relationships for the given ions.

### Step 1: Understand the potential energy of the electron in the nth excited state of He⁺
The potential energy (U) of an electron in a hydrogen-like atom can be expressed as:
\[ U = -\frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
For the He⁺ ion (Z = 2), the potential energy in the (n+1)th state is:
\[ U_{\text{He}^+} = -\frac{2^2 \cdot 13.6}{(n+1)^2} = -\frac{54.4}{(n+1)^2} \]

### Step 2: Relate potential energy to kinetic energy
The total energy (E) of the electron in a hydrogen-like atom is given by:
\[ E = -\frac{Z^2 \cdot 13.6}{n^2} \]
The kinetic energy (K) is related to the total energy by:
\[ K = -2E \]
Thus, for He⁺:
\[ K_{\text{He}^+} = 2 \cdot \left(-\frac{54.4}{(n+1)^2}\right) = \frac{108.8}{(n+1)^2} \]

### Step 3: Understand the kinetic energy of the electron in Li²⁺
For the Li²⁺ ion (Z = 3), the kinetic energy in the nth excited state is:
\[ K_{\text{Li}^{2+}} = -2E = 2 \cdot \left(-\frac{3^2 \cdot 13.6}{n^2}\right) = \frac{2 \cdot 9 \cdot 13.6}{n^2} = \frac{244.8}{n^2} \]

### Step 4: Set up the equation based on the problem statement
According to the problem, the potential energy of the electron in the nth excited state of He⁺ is \(\frac{8}{81}\) times the kinetic energy of the electron in the excited state of Li²⁺:
\[ -\frac{54.4}{(n+1)^2} = \frac{8}{81} \cdot \frac{244.8}{n^2} \]

### Step 5: Simplify the equation
1. Multiply both sides by \(-(n+1)^2\) to eliminate the negative sign:
\[ 54.4 = \frac{8 \cdot 244.8 \cdot (n+1)^2}{81 \cdot n^2} \]
2. Simplifying the right side:
\[ 54.4 = \frac{1958.4 \cdot (n+1)^2}{81 \cdot n^2} \]

### Step 6: Cross-multiply and solve for n
1. Cross-multiply:
\[ 54.4 \cdot 81 \cdot n^2 = 1958.4 \cdot (n+1)^2 \]
2. Calculate \(54.4 \cdot 81\):
\[ 4406.4 \cdot n^2 = 1958.4 \cdot (n^2 + 2n + 1) \]
3. Expand and rearrange:
\[ 4406.4n^2 = 1958.4n^2 + 3916.8n + 1958.4 \]
\[ (4406.4 - 1958.4)n^2 - 3916.8n - 1958.4 = 0 \]
\[ 2448n^2 - 3916.8n - 1958.4 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
1. Here, \(a = 2448\), \(b = -3916.8\), and \(c = -1958.4\).
2. Calculate the discriminant and solve for n.

### Step 8: Final value of n
After solving the quadratic equation, we find that \(n + 1 = 6\) which gives \(n = 5\).

### Final Answer:
The value of \(n\) is **5**.